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Sophie [7]
3 years ago
8

Once a scientist has made a hypothesis, what would they typically do next? (2 points)

Physics
1 answer:
ipn [44]3 years ago
4 0

Answer:

b-testing

Explanation:

First would be observation/research. Then the hypothesis. After that you would test your theory, conduct experiments. And finally, your conclusion- what you got from the whole process basically.

Hope this helps.

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A golfer tees off and hits a golf ball at a speed of 31 m/s and at an angle of 35 degrees. How long was the ball in the air? Rou
Slav-nsk [51]

Answer:

3.6 seconds

Explanation:

Given:

y₀ = y = 0 m

v₀ = 31 sin 35° m/s

a = -9.8 m/s²

Find: t

y = y₀ + v₀ t + ½ at²

0 = 0 + (31 sin 35°) t + ½ (-9.8 m/s²) t²

0 = 17.78t − 4.9t²

0 = t (17.78 − 4.9t)

t = 0 or 3.63

Rounded to the nearest tenth, the ball lands after 3.6 seconds.

4 0
3 years ago
Which is true about a surface wave? ANSWER FAST
Anna [14]

Answer:

D

Explanation:

6 0
3 years ago
How can pressure impact the alignment of minerals within the rock
Sloan [31]
Metamorphism occurs because some minerals are stable only under certain conditions of pressure and temperature. When pressure and temperature change, chemical reactions occur to cause the minerals in the rock to change to an assemblage that is stable at the new pressure and temperature conditions.
6 0
2 years ago
Why would a flat sheet of paper and a wad of paper with the same mass not fall through the air at the same rate?
Gelneren [198K]
The flat sheet of paper has more surface area than the crumpled ball
7 0
3 years ago
Read 2 more answers
A particle of charge 2.0 x 10^-8C experiences an upward force of magnitude 4.0 x10^-6 when it is placed in a particular point in
koban [17]

Answer:

a) The electric field at that point is 2.0\times 10^{2} newtons per coulomb.

b) The electric force is 2.0\times 10^{-6} newtons.

Explanation:

a) Let suppose that electric field is uniform, then the following electric field can be applied:

E = \frac{F_{e}}{q} (1)

Where:

E - Electric field, measured in newtons per coulomb.

F_{e} - Electric force, measured in newtons.

q - Electric charge, measured in coulombs.

If we know that F_{e} = 4.0\times 10^{-6}\,N and q = 2.0\times 10^{-8}\,C, then the electric field at that point is:

E = \frac{4.0\times 10^{-6}\,N}{2.0\times 10^{-8}\,C}

E = 2.0\times 10^{2}\,\frac{N}{C}

The electric field at that point is 2.0\times 10^{2} newtons per coulomb.

b) If we know that E = 2.0\times 10^{2}\,\frac{N}{C} and q = 1.0\times 10^{-8}\,C, then the electric force is:

F_{e} = E\cdot q

F_{e} = \left(2.0\times 10^{2}\,\frac{N}{C} \right)\cdot (1.0\times 10^{-8}\,C)

F_{e} = 2.0\times 10^{-6}\,N

The electric force is 2.0\times 10^{-6} newtons.

7 0
3 years ago
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